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Let $E_1$ be the event that the candidate is female, and $E_2$ be the event that the candidate is male. Let $A$ be the event that the candidate gets a distinction in the written test.
Given that two-thirds of the applicants are females, $P(E_1) = \frac{2}{3}$. Since the rest are males, $P(E_2) = 1 - P(E_1) = 1 - \frac{2}{3} = \frac{1}{3}$.
The probability of a male getting a distinction is $P(A|E_2) = 0.4$, and the probability of a female getting a distinction is $P(A|E_1) = 0.35$.
We want to find the probability that a candidate chosen at random will have a distinction in the written test, which is $P(A)$. Using the law of total probability, we have: $$P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2)$$
Substitute the values we have: $$P(A) = (0.35)\left(\frac{2}{3}\right) + (0.4)\left(\frac{1}{3}\right) = \frac{0.7}{3} + \frac{0.4}{3} = \frac{1.1}{3} = \frac{11}{30}$$
Therefore, the probability that the candidate chosen at random will have a distinction in the written test is $\frac{11}{30}$.
Final Answer: 11/30
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